This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Chapter 08.04 RungeKutta 4th Order Method for Ordinary Differential Equations After reading this chapter, you should be able to 1. develop RungeKutta 4th order method for solving ordinary differential equations, 2. find the effect size of step size has on the solution, 3. know the formulas for other versions of the RungeKutta 4th order method What is the RungeKutta 4th order method? RungeKutta 4th order method is a numerical technique used to solve ordinary differential equation of the form dy = f ( x, y ) , y ( 0 ) = y 0 dx So only first order ordinary differential equations can be solved by using the RungeKutta 4 th order method. In other sections, we have discussed how Euler and RungeKutta methods are used to solve higher order ordinary differential equations or coupled (simultaneous) differential equations. How does one write a first order differential equation in the above form? Example 1 Rewrite dy + 2 y = 1.3e − x , y ( 0 ) = 5 dx in dy = f ( x, y ), y (0) = y 0 form. dx 08.04.1 08.04.2 Solution dy + 2 y = 1.3e − x , y ( 0 ) = 5 dx dy = 1.3e − x − 2 y, y ( 0 ) = 5 dx In this case f ( x, y ) = 1.3e − x − 2 y Example 2 Rewrite ey in dy = f ( x, y ), y (0) = y 0 form. dx Solution dy + x 2 y 2 = 2 sin(3 x), y ( 0 ) = 5 dx Chapter 08.04 dy + x 2 y 2 = 2 sin(3 x), y ( 0 ) = 5 dx dy 2 sin(3 x) − x 2 y 2 = , y( 0) = 5 dx ey In this case 2 sin(3 x) − x 2 y 2 f ( x, y ) = ey The RungeKutta 4th order method is based on the following yi +1 = yi + ( a1k1 + a2 k 2 + a3 k 3 + a4 k 4 ) h where knowing the value of y = y i at xi , we can find the value of y = yi +1 at xi +1 , and h = xi +1 − xi Equation (1) is equated to the first five terms of Taylor series 2 3 3 dy yi +1 = yi + ( xi +1 − xi ) + 1 d y xi , yi ( xi +1 − xi ) 2 + 1 d y xi , yi ( xi +1 − xi ) xi , yi 2 3 dx 2! dx 3! dx 4 1d y ( xi +1 − xi ) 4 + 4 xi , yi 4! dx dy = f ( x, y ) and xi +1 − xi = h Knowing that dx 1 1 1 y i +1 = y i + f ( xi , y i ) h + f ' ( xi , y i ) h 2 + f '' ( xi , y i ) h 3 + f ''' ( xi , y i ) h 4 2! 3! 4! (3) Based on equating Equation (2) and Equation (3), one of the popular solutions used is ey (1) (2) RungeKutta 4th Order Method 1 ( k1 + 2 k 2 + 2k 3 + k 4 ) h 6 k1 = f ( x i , y i ) y i +1 = y i + 1 1 k 2 = f xi + h, yi + k1h 2 2 1 1 k 3 = f xi + h, y i + k 2 h 2 2 k 4 = f ( xi + h, y i + k 3 h ) Example 3 08.04.3 (4) (5a) (5b) (5c) (5d) A ball at 1200 K is allowed to cool down in air at an ambient temperature of 300 K. Assuming heat is lost only due to radiation, the differential equation for the temperature of the ball is given by dθ = −2.2067 ×10 −12 θ 4 − 81× 108 , θ ( 0 ) = 1200 K dt where θ is in K and t in seconds. Find the temperature at t = 480 seconds using RungeKutta 4th order method. Assume a step size of h = 240 seconds. Solution ( ) dθ = −2.2067 × 10 −12 θ 4 − 81 × 10 8 dt f ( t , θ ) = −2.2067 × 10 −12 θ 4 − 81 × 10 8 1 θ i +1 = θ i + ( k1 + 2k 2 + 2k 3 + k 4 ) h 6 For i = 0 , t 0 = 0 , θ 0 = 1200K k1 = f ( t0 , θ 0 ) = f ( 0,1200 ) = −2.2067 × 10 −12 1200 4 − 81 × 10 8 = −4.5579 1 1 k 2 = f t0 + h, θ 0 + k1h 2 2 1 1 = f 0 + ( 240) ,1200 + ( − 4.5579 ) × 240 2 2 = f (120,653.05) = −2.2067 × 10 −12 653.05 4 − 81 × 10 8 = −0.38347 1 1 k 3 = f t 0 + h, θ 0 + k 2 h 2 2 ( ( ) ) ( ) ( ) 08.04.4 1 1 = f 0 + ( 240) ,1200 + ( − 0.38347 ) × 240 2 2 = f (120,1154.0 ) = −2.2067 ×10 −12 1154.0 4 − 81×108 = −3.8954 k 4 = f ( t0 + h, θ 0 + k3h ) = f ( 0 + 240,1200 + ( − 3.894 ) × 240) = f ( 240,265.10) = −2.2067 ×10 −12 265.10 4 − 81×108 = 0.0069750 1 θ 1 = θ 0 + ( k1 + 2 k 2 + 2 k 3 + k 4 ) h 6 1 = 1200 + ( − 4.5579 + 2( − 0.38347 ) + 2( − 3.8954) + ( 0.069750 ) ) 240 6 = 1200 + ( − 2.1848) × 240 = 675.65 K θ1 is the approximate temperature at t = t1 = t0 + h = 0 + 240 = 240 θ1 = θ ( 240 ) ≈ 675.65 K For i = 1, t1 = 240, θ1 = 675.65 K k1 = f ( t1 , θ 1 ) = f ( 240,675.65) = −2.2067 × 10 −12 675.65 4 − 81 × 10 8 = −0.44199 1 1 k 2 = f t1 + h, θ1 + k1h 2 2 1 1 = f 240 + ( 240) ,675.65 + ( − 0.44199) 240 2 2 = f ( 360,622.61) = −2.2067 × 10−12 622.614 − 81 × 108 = −0.31372 1 1 k3 = f t1 + h, θ1 + k 2 h 2 2 Chapter 08.04 ( ) ( ) ( ) ( ) RungeKutta 4th Order Method 08.04.5 1 1 = f 240 + ( 240 ) ,675.65 + ( − 0.31372 ) × 240 2 2 = f ( 360,638.00) = −2.2067 × 10 −12 638.00 4 − 81 × 10 8 = −0.34775 k 4 = f ( t1 + h,θ1 + k3h ) = f ( 240 + 240,675.65 + ( − 0.34775) × 240) = f ( 480,592.19 ) = 2.2067 × 10 −12 592.19 4 − 81 × 10 8 = −0.25351 1 θ 2 = θ1 + (k1 + 2k 2 + 2k 3 + k 4 )h 6 1 = 675.65 + ( − 0.44199 + 2( − 0.31372 ) + 2( − 0.34775) + ( − 0.25351) ) × 240 6 1 = 675.65 + ( − 2.0184 ) × 240 6 = 594.91 K θ 2 is the approximate temperature at t = t2 = t1 + h = 240 + 240 = 480 ( ) ( ) θ 2 = θ ( 480) ≈ 594.91 K Figure 1 compares the exact solution with the numerical solution using the RungeKutta 4th order method with different step sizes. 08.04.6 Chapter 08.04 1600 Temperature, θ(K) 1200 800 400 0 0 400 Time,t(sec) 200 h=480 400 600 h=120 Exact h=240 Figure 1 Comparison of RungeKutta 4th order method with exact solution for different step sizes. Table 1 and Figure 2 show the effect of step size on the value of the calculated temperature at t = 480 seconds. Table 1 Value of temperature at time, t = 480 s for different step sizes Step size, h
800480 Temperature, θ(480) 240 600120 60 30 400
200 0 0 200 100 θ ( 480) 90.278 594.91 646.16 647.54 647.57 Et 737.85 52.660 1.4122 0.033626 0.0008690 0  εt  % 113.94 8.1319 0.21807 0.0051926 0.00013419 200 300 400 500 Step size, h Figure 2 Effect of step size in RungeKutta 4th order method. In Figure 3, we are comparing the exact results with Euler’s method (RungeKutta 1st order method), Heun’s method (RungeKutta 2nd order method), and RungeKutta 4th order method. RungeKutta 4th Order Method 08.04.7 The formula described in this chapter was developed by Runge. This formula is same as Simpson’s 1/3 rule, if f ( x, y ) were only a function of x . There are other versions of the 4th order method just like there are several versions of the second order methods. The formula developed by Kutta is 1 y i +1 = y i + ( k1 + 3k 2 + 3k 3 + k 4 ) h (6) 8 where k1 = f ( xi , y i ) (7a) 1 1 k 2 = f xi + h, y i + hk1 3 3 2 1 k 3 = f xi + h, y i − hk1 + hk 2 3 3 k 4 = f ( xi + h, y i + hk1 − hk 2 + hk 3 ) (7b) (7c) (7d) This formula is the same as the Simpson’s 3/8 rule, if f ( x, y ) is only a function of x .
1400 Temperature, θ(K) 1200 1000 800 600 400 200 0 4th order Exact Heun Euler 0 100 200 300 400 500 Time, t(sec) Figure 3 Comparison of RungeKutta methods of 1st (Euler), 2nd, and 4th order. ORDINARY DIFFERENTIAL EQUATIONS Topic RungeKutta 4th order method Summary Textbook notes on the RungeKutta 4th order method for solving ordinary differential equations. Major General Engineering Authors Autar Kaw Last Revised May 1, 2011 Web Site http://numericalmethods.eng.usf.edu ...
View
Full
Document
This note was uploaded on 05/01/2011 for the course CHBE 2120 taught by Professor Gallivan during the Spring '07 term at Georgia Institute of Technology.
 Spring '07
 Gallivan

Click to edit the document details